Question

Assuming complete dissociation, what is the pH of a 4.47 mg/L4.47 mg/L Ba(OH)2Ba(OH)2 solution?

pH=

Answer 1

Consider 1 L of solution.

Then mass of Ba(OH)2 = 4.47 mg

Lets find the concentration of Ba(OH)2 solution

Molar mass of Ba(OH)2,

MM = 1*MM(Ba) + 2*MM(O) + 2*MM(H)

= 1*137.3 + 2*16.0 + 2*1.008

= 171.316 g/mol

mass(Ba(OH)2)= 4.47 mg

= 0.00447 g

use:

number of mol of Ba(OH)2,

n = mass of Ba(OH)2/molar mass of Ba(OH)2

=(4.47*10^-3 g)/(1.713*10^2 g/mol)

= 2.609*10^-5 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 2.609*10^-5/1

= 2.609*10^-5 M

So,

[OH-] = 2*[Ba(OH)2]

= 2*2.609*10^-5 M

= 5.218*10^-5 M

use:

pOH = -log [OH-]

= -log (5.218*10^-5)

= 4.2825

use:

PH = 14 - pOH

= 14 - 4.2825

= 9.7175

Answer: 9.72